Keep all ten seats when people are absent

Order is irrelevant: $\binom{10}4=210$.

A four-person committee contains at most two complete couples. The forbidden committees are exactly two of the three couples: $\binom32=3$. Thus $210-3=207$. Selecting one member from each of different couples does not create a complete couple.

The printed solution uses 7 men and 3 women. All-women committees of size 4 are impossible. The only forbidden case is four men, $\binom74=35$, giving $210-35=175$.

Treat each couple as one block, alongside the 4 single men:7 distinguishable circular objects. Use one single man as the reference. Each couple has $2!$ internal orders, so $(7-1)!\,(2!)^3=5760$.

There are still 10 seats, now containing 9 distinct people and one empty position. Use that unique vacancy as the circular reference: $(10-1)!=9!=362880$. The identity of the absent man is fixed, so no factor for choosing him is needed.

With a particular couple absent, arrange 8 distinct people and 2 indistinguishable empty positions around the 10-seat circle. Fix a remaining single man as reference, then divide by $2!$ for the vacancies: $(10-1)!/2!=181440$. The source treats rotations as equivalent because the seats are not numbered, unlike Q8(ii).